A block of mass 2 kg is moving with a speed of $10 m s^{- 1}$ on a rough horizontal floor till it stops. Determine the work done against the friction.

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Solution

Step 1: Given
The initial velocity of the block is $10 m s^{- 1}$
The final velocity of the block is $0 m s^{- 1}$
The weight of the block is 2 kg.
Step 2: Formula & solution
Work done by friction will be equal to the change in kinetic energy.
$∆ K E = w o r k d o n e f r i c t i o n t o s t o p t h e b o d y \frac{1}{2} \times 2 \times {(0)}^{2} - \frac{1}{2} \times 2 \times {(10)}^{2} = W_{f r i c t i o n} - 100 J = W_{f r i c t i o n}$
Here, the - sign indicates that the work done is in the direction opposite to that of motion.
Work done by friction force is equal to 100 J.

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A block of mass 2 kg is moving with a speed of 10ms-1 on a rough horizontal floor till it stops. Determine the work done against the friction.

A block of mass 2 kg is moving with a speed of $10 m s^{- 1}$ on a rough horizontal floor till it stops. Determine the work done against the friction.